The definition of proper morphism by separated, finite type, universally closed properties doesn’t look enlightening to me. I am trying to look for a simpler characterization for this concept.

As we know properness in algebraic geometry is the analog notion of compactness in topology. In topology the definition of the proper map is that the inverse image of every compact subspace is a compact subspace, I am wondering is there an analog definition for proper morphisms between schemes in algebraic geometry.

A natural statement one might wish to hold is the following:

Let S be a scheme. A morphism between S schemes is proper if and only if the inverse image of every proper S subscheme is a proper S subscheme.

Here we put the reduced induced subscheme structure on the inverse image of a closed subscheme. Is this statement true? If not, is it true in some special cases, say k schemes, where k is a field? Or are there some related statements that hold? Thank you!


1 Answer 1


You are correct that if $f : X \to Y$ is proper over some base $S$ then it has the property you described. If $Z \subset Y$ is a closed subscheme proper over $S$ then $f^{-1}(Z) \subset X$ is a closed subscheme proper over $S$.

Because $f^{-1}(Z)$ has $f^{-1}(Z)$ has a natural subscheme structure as $f^{-1}(Z) = Z \times_Y X$ which is not always reduced, it is probably better to require that $f^{-1}(Z)$ be proper with this standard subscheme structure.

However, it's also true if we give $f^{-1}(Z)$ the reduced subscheme structure because this is a closed subscheme of $f^{-1}(Z)$ with its standard structure and closed subschemes of proper schemes are proper.

Now let me prove the claim. The map $f^{-1}(Z) \to Z$ is proper because it is the base change of $f : X \to Y$. Therefore the composition $f^{-1}(Z) \to Z \to S$ is proper showing that $f^{-1}(Z)$ is proper over $S$.

However, I do not believe this property is enough to characterize proper morphisms even for varieties over an algebraically closed field. The problem is ``there aren't enough proper closed subvarieties''. For example, $\mathbb{A}^1 \setminus \{ 0 \} \to \mathbb{A}^1$ is not proper but the only proper closed subvarities of $\mathbb{A}^1$ are points whose preimages are also points.

In fact, we can make examples where $f : X \to Y$ is a closed surjective map of varieties over $\mathbb{C}$ with your property but not proper. For example, let $Y$ be the affine nodal curve $y^2 = x^2(x-1)$ and $X = \mathbb{A}^1 \setminus \{ i \}$. The normalization map $\nu : \tilde{Y} \to Y$ where $\tilde{Y} = \mathbb{A}^1$ sends $t \mapsto (t^2 + 1, t(t^2 + 1))$ so take $X \to \tilde{Y} \to Y$ i.e. we remove one of the two preimages of the node. Then $f$ is closed and surjective (even bijective!) pulling back proper closed subvarities (points) to proper closed subvarities. However, $f$ is not proper. To see this, consider $\tilde{f} : X \times \tilde{Y} \to Y \times \tilde{Y}$ and the closed set $$\Delta = \{ (x,x) \mid x \in X \} \subset X \times \tilde{Y} $$ however $\tilde{f}(\Delta) = \{ (f(x), x) \mid x \in X \} \subset Y \times \tilde{Y}$ is the graph of $f$ minus one point and thus not closed.

However, for smooth varieties over $\mathbb{C}$, we can view an algebraic map $f : X \to Y$ as a map of complex manifolds $f^{\mathrm{an}} : X^{\mathrm{an}} \to Y^{\mathrm{an}}$ viewing $X, Y$ as complex manifolds with the analytic topology. Then it is a comforting fact that $f$ is proper iff $f^{\mathrm{an}}$ is proper in the usual topology sense.

  • 2
    $\begingroup$ +1, a good answer. One small comment - the reduced induced structure on $f^{-1}(Z)$ expresses that as a closed subscheme of $f^{-1}(Z)=Z\times_Y X$ with the usual subscheme structure, so $f^{-1}(Z)$ with the reduced induced subscheme structure is proper over $S$ too. Therefore you can upgrade "I think this is true..." to "it's also true that..." in the parenthetical. (I agree with you that the subscheme structure given by $Z\times_YX$ is the more natural one to use.) $\endgroup$
    – KReiser
    Jan 22, 2021 at 21:09
  • $\begingroup$ Thank you @KReiser, I clearly didn't give it enough thought! $\endgroup$
    – Ben C
    Jan 22, 2021 at 21:32
  • $\begingroup$ @Ben C Thank you for the answer. The counterexamples you provided illustrate the failure of the other indication well. I have a clearer picture of the definitions in the two settings now. $\endgroup$
    – user875280
    Jan 23, 2021 at 5:57

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